I've been assigned to write a program which then calculates the Laurent series of a function using Python. I've already found a library called SymPy for symbolical computations in Python but the problem is I have no idea how should I produce the Laurent series in the program. Of course I'm familiar with the concept, but I've always calculated the Laurent series in an ad hoc way
我被分配编写一个程序，然后使用Python计算一个函数的Laurent系列。 我已经找到了一个名为SymPy的库，用于Python中的符号计算，但问题是我不知道该如何在程序中生成Laurent系列。 当然，我对这个概念很熟悉，但我一直使用泰勒级数以特殊方式计算Laurent系列，从未使用算法方法。 如果有人帮我找到使用下面问题中提到的输入来生成Laurent系列的算法，我将不胜感激:-) Given a fractional function containing polynomials in b
I want to get a list of the column headers from a pandas DataFrame. The DataFrame will come from user input so I won't know how many columns there will be or what they will be called. For example, if I'm given a DataFrame like this: >>> my_dataframe y gdp cap 0 1 2 5 1 2 3 9 2 8 7 2 3 3 4 7 4 6 7 7 5 4 8 3 6 8 2 8
我想从pandas DataFrame中获取列标题的列表。 DataFrame将来自用户输入，所以我不知道将有多少列或将被调用。 例如，如果我这样给一个DataFrame： >>> my_dataframe y gdp cap 0 1 2 5 1 2 3 9 2 8 7 2 3 3 4 7 4 6 7 7 5 4 8 3 6 8 2 8 7 9 9 10 8 6 6 4 9 10 10 7 我想要得到这样的列表： >>> header_list [y, gd
I have a DataFrames from pandas: import pandas as pd inp = [{'c1':10, 'c2':100}, {'c1':11,'c2':110}, {'c1':12,'c2':120}] df = pd.DataFrame(inp) print df Output: c1 c2 0 10 100 1 11 110 2 12 120 Now I want to iterate over the rows of the above frame. For every row I want to be able to access its elements (values in cells) by the name of the columns. So, for example, I would like t
我有熊猫的DataFrames： import pandas as pd inp = [{'c1':10, 'c2':100}, {'c1':11,'c2':110}, {'c1':12,'c2':120}] df = pd.DataFrame(inp) print df 输出： c1 c2 0 10 100 1 11 110 2 12 120 现在我想遍历上述帧的行。 对于每一行，我希望能够通过列的名称访问其元素（单元格中的值）。 所以，例如，我想有这样的事情： for row in df.rows: print row['c1'], row['c2'] 熊猫可以这样做吗？ 我发现
I have the following indexed DataFrame with named columns and rows not- continuous numbers: a b c d 2 0.671399 0.101208 -0.181532 0.241273 3 0.446172 -0.243316 0.051767 1.577318 5 0.614758 0.075793 -0.451460 -0.012493 I would like to add a new column, 'e' , to the existing data frame and do not want to change anything in the data frame (ie, the
我有以下索引DataFrame与命名的列和行不连续的数字： a b c d 2 0.671399 0.101208 -0.181532 0.241273 3 0.446172 -0.243316 0.051767 1.577318 5 0.614758 0.075793 -0.451460 -0.012493 我想在现有的数据框中添加一个新的列'e' ，并且不想在数据框中更改任何内容（即新列的长度始终与DataFrame的长度相同）。 0 -0.335485 1 -1.166658 2 -0.385571 dtype: float64
I have a list of dicts: b = [{u'TOT_PTS_Misc': u'Utley, Alex', u'Total_Points': 96.0}, {u'TOT_PTS_Misc': u'Russo, Brandon', u'Total_Points': 96.0}, {u'TOT_PTS_Misc': u'Chappell, Justin', u'Total_Points': 96.0}, {u'TOT_PTS_Misc': u'Foster, Toney', u'Total_Points': 80.0}, {u'TOT_PTS_Misc': u'Lawson, Roman', u'Total_Points': 80.0}, {u'TOT_PTS_Misc': u'Lempke, Sam', u'Total_Points': 80.0}, {u'
我有一个列表： b = [{u'TOT_PTS_Misc': u'Utley, Alex', u'Total_Points': 96.0}, {u'TOT_PTS_Misc': u'Russo, Brandon', u'Total_Points': 96.0}, {u'TOT_PTS_Misc': u'Chappell, Justin', u'Total_Points': 96.0}, {u'TOT_PTS_Misc': u'Foster, Toney', u'Total_Points': 80.0}, {u'TOT_PTS_Misc': u'Lawson, Roman', u'Total_Points': 80.0}, {u'TOT_PTS_Misc': u'Lempke, Sam', u'Total_Points': 80.0}, {u'TOT_PTS_M
I have the following list created from a sorted csv list1 = sorted(csv1, key=operator.itemgetter(1)) I would actually like to sort the list by two criteria: first by the value in field 1 and then by the value in field 2. How do I do this? 喜欢这个： import operator list1 = sorted(csv1, key=operator.itemgetter(1, 2)) Replying to this dead thread for archive. No need to import anything when u
我有一个从排序的CSV创建的以下列表 list1 = sorted(csv1, key=operator.itemgetter(1)) 我实际上希望按照两个标准对列表进行排序：首先按字段1中的值，然后按字段2中的值排序。我该怎么做？ 喜欢这个： import operator list1 = sorted(csv1, key=operator.itemgetter(1, 2)) 回复这个死去的线程进行归档。 使用lambda函数时无需导入任何东西。 下面的排序list的第一个元素，然后通过第二个元素。 sorted(list, key=lam
This question already has an answer here: How to pair socks from a pile efficiently? 36 answers Python functions with multiple parameter brackets 3 answers Syntax of Keras Functional API 2 answers There are multiple ways that this code can work, 2 come into my mind right now: Using the built in __call__ function of classes >>> class MyClass: ... def __init__(self, name):
这个问题在这里已经有了答案： 如何有效地从一堆袜子中配对？ 36个答案 Python函数带有多个参数括号3个答案 Keras Functional API 2的语法的答案 这个代码可以有多种工作方式，现在我想起来了： 使用内置的__call__函数的类 >>> class MyClass: ... def __init__(self, name): ... self.name = name ... def __call__(self, word): ... print(self.name, 'says', word) ...
Is there a reason to prefer using map() over list comprehension or vice versa? Is either of them generally more efficient or considered generally more pythonic than the other? map may be microscopically faster in some cases (when you're NOT making a lambda for the purpose, but using the same function in map and a listcomp). List comprehensions may be faster in other cases and most (not al
有理由更喜欢使用map()不是列表理解，反之亦然？ 他们中的任何一个通常更有效率或被认为通常比另一个更pythonic？ 在某些情况下， map可能会在显微镜下更快（当您不为此目的制作lambda时，但在地图和listcomp中使用相同的功能时）。 在其他情况下，列表理解可能会更快，大多数（并非全部）pythonistas认为它们更直接，更清晰。 当使用完全相同的功能时，地图的速度优势很小的一个例子： $ python -mtimeit -s'xs=range(10)
I coded a solution for DFS non-recursive, but i can't modify it to make a topological sort: def dfs(graph,start): path = [] stack = [start] while stack != []: v = stack.pop() if v not in path: path.append(v) for w in reversed(graph[v]): if w not in path and not w in stack: stack.append(w) return path Any ideas how to
我编写了一个DFS非递归解决方案，但我无法对其进行修改以进行拓扑排序： def dfs(graph,start): path = [] stack = [start] while stack != []: v = stack.pop() if v not in path: path.append(v) for w in reversed(graph[v]): if w not in path and not w in stack: stack.append(w) return path 任何想法如何修改它？ 随着递归版本，我可以
I am configuring access control for a web application based on the Pyramid framework. I am setting up permissions for my view callables using the @view_config decorator. I have two permissions, namely 'read' and 'write' . Now, I want certain views to require both permissions. I was unable to figure out how to do this with view_config - am I missing something, or is there maybe
我正在为基于Pyramid框架的Web应用程序配置访问控制。 我正在使用@view_config修饰器为我的视图@view_config设置权限。 我有两个权限，分别是'read'和'write' 。 现在，我想要某些视图需要这两个权限。 我无法弄清楚如何使用view_config来做到这view_config - 我是否错过了一些东西，或者有另一种方法可以做到这一点？ 设置readwrite权限。 每个视图只获得一个权限，但每个主体可以映射到许多权限。